# Find the equation of a circle, if the end points of the diameters are A (2, -3) and B(-3, 5) ?

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Find the equation of a circle, if the end points of the diameters are A (2, -3) and B(-3, 5) ?
1. x2 + y2 - x - 2y - 21 = 0
2. x2 + y2 + x - 2y + 21 = 0
3. x2 + y2 + x + 2y - 21 = 0
4. x2 + y2 + x - 2y - 21 = 0
5. None of these

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Correct Answer - Option 4 : x2 + y2 + x - 2y - 21 = 0

CONCEPT:

If (x1, y1) and (x2, y2) are the end points of the diameter of a circle. Then the equation of such a circle is (x – x1) ⋅ (x – x2) + (y – y1) (y – y2) = 0

CALCULATION:

Given: The end points of the diameter of a circle are  A (2, - 3) and B(- 3, 5)

As we know that, if (x1, y1) and (x2, y2) are the end points of the diameter of a circle then the equation of such a circle is (x – x1) ⋅ (x – x2) + (y – y1) (y – y2) = 0

Here, x1 = 2, y1 = -3, x2 = -3 and y2 = 5

⇒ (x - 2) ⋅ (x + 3) + (y + 3) ⋅ (y - 5) = 0

⇒ x2 + y2 + x - 2y - 21 = 0

So, the equation of the required circle is: x2 + y2 + x - 2y - 21 = 0

Hence, option D is the correct answer.